A circle is inscribed in a regular hexagon, a smaller regular hexagon is inscribed in the circle. Compute the ratio of the area of the larger hexagon to the area of the smaller hexagon.

I have no idea how to solve this problem. I would be much obliged if anyone were to help me.

Mar 28th 2011, 01:12 PM

TheChaz

Draw a picture. Even a rough sketch should suffice.
Connect the center to a vertex of the bigger hexagon.
Connect the center to a nearby vertex of the smaller hexagon (which will be the midpoint of a side of the larger...).

Use some angle and side knowledge.
Feel free to ask for more hints!

Mar 28th 2011, 01:32 PM

NOX Andrew

Thank you for your help. I drew the figures as you suggested, but I don't understand how one would know a line drawn from the center of the circle to a nearby vertex of the smaller hexagon passes through the midpoint of a side of the larger hexagon.

Mar 28th 2011, 01:57 PM

TheChaz

My phrasing was a little misleading...
It doesn't necessarily pass through the midpoint of the larger side, but you can rotate it without affecting anything else.

Mar 28th 2011, 05:25 PM

Plato

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Quote:

Originally Posted by NOX Andrew

A circle is inscribed in a regular hexagon, a smaller regular hexagon is inscribed in the circle. Compute the ratio of the area of the larger hexagon to the area of the smaller hexagon.
I have no idea how to solve this problem. I would be much obliged if anyone were to help me.

Look at the diagram. Can you calculate the area of the part in yellow?

Mar 28th 2011, 05:43 PM

TheChaz

Sure wish I knew how to make a diagram! It's very simple, once you see the 30-60-90 triangle(s). Maybe Plato will upload a file suggestive of this method...

Mar 28th 2011, 05:50 PM

Wilmer

Or go like this:
1: draw the larger hexagon
2: draw its inscribed circle : notice that the 6 tangent points are located at midpoint of the hexagon's sides
3: draw the smaller hexagon by joining these 6 points
Howz dat?
Then I suggest you assign 1 as length of circle's radius; agree Chaz ?

EDIT:
whoops...didn't see Plato's diagram...
mine will look different, and involve 6 congruent triangles as the difference in areas...

Mar 28th 2011, 06:16 PM

TheChaz

I would use "1" for the length of the longer... (can't remember the vocab word) center-vertex segment.
This will be the hypotenuse. 1/2 will be the shorter leg... the longer leg is the center-vertex of the smaller pentagon. So this boils down to the ratio of sides of a 30-60-90 triangle (or a trig function).

Using 1 for the radius will make it look more like the unit circle ;)

Mar 28th 2011, 06:53 PM

Wilmer

Quote:

Originally Posted by TheChaz

I would use "1" for the length of the longer... (can't remember the vocab word) center-vertex segment.
This will be the hypotenuse. 1/2 will be the shorter leg... the longer leg is the center-vertex of the smaller pentagon. So this boils down to the ratio of sides of a 30-60-90 triangle (or a trig function).
Using 1 for the radius will make it look more like the unit circle ;)

Hello NOX Andrew,
Drawone sector of Plato's diagram.There are two overlapping equilateral triangles.If the smaller has a side length of 1 its apothem is 1/2*rad3. The larger has an apothem of one and a side lengthof2/rad3.